# About a type of limit

• B

## Main Question or Discussion Point

Would this $f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0$ hold if $\frac{f(x)}{g(x)}\neq c$?

Math_QED
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It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an $\implies$ and then an 'if' after the implies.

What is $a$ ? What is $c$? Where are $f,g$ defined?

RPinPA
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Would this $f'(x)>g'(x)\,\forall x\in [a,\infty)\text{ and }f,\,g\underset{\infty}{\to}0\Rightarrow \lim_{x\to\infty}f(x)/g(x)=0$ hold if $\frac{f(x)}{g(x)}\neq c$?
This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some $a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)$ and $f,\,g\underset{\infty}{\to}0$ (Does this mean $\lim_{x\to\infty}f, g = 0$?) then $\lim_{x\to\infty}f(x)/g(x)=0$. Does the theorem hold if $\frac{f(x)}{g(x)}\neq c$?"

Or are you asking if there is such a theorem?

WWGD
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If I understood correctly, $1/x ,1/(x+1)$ are counters.

It is unclear to me what you mean. Can you be a little more careful describe what you mean? You have an $\implies$ and then an 'if' after the implies.

What is $a$ ? What is $c$? Where are $f,g$ defined?
This is a little cryptic. Here's what I think you are saying:

"There is a theorem that if for some $a \in \mathbb R, f'(x)>g'(x)\,\forall x\in [a,\infty)$ and $f,\,g\underset{\infty}{\to}0$ (Does this mean $\lim_{x\to\infty}f, g = 0$?) then $\lim_{x\to\infty}f(x)/g(x)=0$. Does the theorem hold if $\frac{f(x)}{g(x)}\neq c$?"

Or are you asking if there is such a theorem?
What I want to say is, if there're two functions $f(x)$ and $g(x)$ who are going to $0$ as $x\to\infty$, and such that after a certain $x\,(=a)$ we have $f'(x)>g'(x)$ then $\frac{f(x)}{g(x)} \underset{\infty}{\to} 0$ provided that $f(x)\neq cg(x)$ where $c\in\mathbb{R}$, elsewise the limit would equal $c$.
I realised some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if $f$ goes to $0$ faster than $g$ then ..., the correct way to translate this in math is $|f'(x)|<|g'(x)|$ (provided that both $f$ and $g$ are going to $0$).

If I understood correctly, $1/x ,1/(x+1)$ are counters.
No, they don't come under this since they have the same rate of change.

WWGD
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No, they don't come under this since they have the same rate of change.
Respective derivatives are $-1/x^2 >-1/(x+1)^2$

Respective derivatives are $-1/x^2 >-1/(x+1)^2$
Right sorry. I think, then, that I can say something about the convergence of the limit given the assumptions? And then it follows that it is $0$ if the functions are not equivalent at $\infty$?

PeroK
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What I want to say is, if there're two functions $f(x)$ and $g(x)$ who are going to $0$ as $x\to\infty$, and such that after a certain $x\,(=a)$ we have $f'(x)>g'(x)$ then $\frac{f(x)}{g(x)} \underset{\infty}{\to} 0$ provided that $f(x)\neq cg(x)$ where $c\in\mathbb{R}$, elsewise the limit would equal $c$.
I realised some holes in this through math.stackexchange. I was initially thinking of two decreasing functions, i.e both positive.
However, since what I want to say is if $f$ goes to $0$ faster than $g$ then ..., the correct way to translate this in math is $|f'(x)|<|g'(x)|$ (provided that both $f$ and $g$ are going to $0$).
I can't see any reason why this would be true.

I can't see any reason why this would be true.
$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use $O(10^{-n})$? Not sure, I need to study big-O notation.

PeroK
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$$N=n-m>0,\,\,\pm\frac{10^{-n}}{10^{-m}}=\pm10^{-n+m}=\pm10^{-N}=\pm\frac{1}{e^{N\ln10}}\underset{N\to \infty}{\to}0$$
By $10^{k}$ I just wanted to represent a kind of manual evaluation of a limit where the nominator is going to zero faster than the denominator. Maybe I should use $O(10^{-n})$? Not sure, I need to study big-O notation.
I can't see the relationship between that and what you posted before.

I can't see the relationship between that and what you posted before.
Well, if $|f'(x)|<|g'(x)|$ and they are both going to $0$, it means that $f$ is going to $0$ faster, and so the limit should be $0$.

PeroK
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Well, if $|f'(x)|<|g'(x)|$ and they are both going to $0$, it means that $f$ is going to $0$ faster, and so the limit should be $0$.
That relationship means that $g(x)$ is changing more quickly than $f(x)$.

You've got that the wrong way round.

PeroK
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Well, if $|f'(x)|<|g'(x)|$ and they are both going to $0$, it means that $f$ is going to $0$ faster, and so the limit should be $0$.
In any case, you already have the counterexample: $f(x) = \frac 1 2 g(x)$.

Even if you exclude that precise counterexample, you only need two functions where $f(x) \approx \frac 1 2 g(x)$. For example:

$f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)$

That relationship means that $g(x)$ is changing more quickly than $f(x)$.

You've got that the wrong way round.
If $f$ is going faster to $0$ than $g$, then $|df|$ would be less than $|dg|$, because it is already "at zero" so there isn't much to change.
In any case, you already have the counterexample: $f(x) = \frac 1 2 g(x)$.

Even if you exclude that precise counterexample, you only need two functions where $f(x) \approx \frac 1 2 g(x)$. For example:

$f(x) = (\frac 1 2 + \frac{\sin x}{x})g(x)$
Right, see post #8.
through the different counter-examples given, it is clear that this is not true. however, given the premises, it seems that the limit always converges (maybe), and in the special case that $f\underset{\infty}{\not\sim}g$, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.

PeroK
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however, given the premises, it seems that the limit always converges (maybe), and in the special case that $f\underset{\infty}{\not\sim}g$, it is zero (maybe). This is just an intuitive exercise, I just feel like this might work out.
Not that either. Just amend my last function:

$f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)$

Then the limit $f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}$ is undefined.

• archaic and Math_QED
Not that either. Just amend my last function:

$f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)$

Then the limit $f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}$ is undefined.
wops

Not that either. Just amend my last function:

$f(x) = (\frac 1 2 + \frac{\sin x}{10})g(x)$

Then the limit $f(x)/g(x) = \frac 1 2 + \frac{\sin x}{10}$ is undefined.
It doesn't satisfy $|f'(x)|<|g'(x)|$ so it doesn't come under this.

PeroK
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wops
Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty$.

Any ideas?

PeroK
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It doesn't satisfy $|f'(x)|<|g'(x)|$ so it doesn't come under this.
With $g(x) = e^{-x}$ it all works out.

• archaic
Seriously, you ought to think about how I created these counterexamples. They only really involved some elementary geometrical thinking.

The only one we don't have is where $\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = +\infty$.

Any ideas?
you mean given the premises?

PeroK
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you mean given the premises?
Yes.

PeroK
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you mean given the premises?
Maybe this is the result that could be proved? If $|f'(x)| < |g'(x)|$, then $f(x)$ is changing less than $g(x)$, so $f(x)/g(x)$ cannot become arbitrarily large. So:

$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty$?

Maybe this is the result that could be proved? If $|f'(x)| < |g'(x)|$, then $f(x)$ is changing less than $g(x)$, so $f(x)/g(x)$ cannot become arbitrarily large. So:

$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} \ne +\infty$?
I thought you meant that there was an example where that limit goes to infinity...
But yes, it seems reasonable that given those premises the limit can't be arbitrarily large.

PeroK